1991 AMC 8 真题与在线限时测验

计时

40:00

1.

下列表达式的值是多少？ $\begin{aligned} &1{,}000{,}000{,}000{,}000 \\ &\quad {}- 777{,}777{,}777{,}777 \end{aligned}\text{？}$

What is the value of $\begin{aligned} &1{,}000{,}000{,}000{,}000 \\ &\quad {}- 777{,}777{,}777{,}777? \end{aligned}$

$222{,}222{,}222{,}222$

$222{,}222{,}222{,}223$

$233{,}333{,}333{,}333$

$322{,}222{,}222{,}223$

$333{,}333{,}333{,}333$

答案：B

知识点：位值

难度评级：450

小提示：

想一想要给 $777{,}777{,}777{,}777$ 加多少，才能到达 $1{,}000{,}000{,}000{,}000$

Think of the amount you must add to $777{,}777{,}777{,}777$ to reach $1{,}000{,}000{,}000{,}000$

大提示：

从右往左看：个位要加到 $10$ 并进位，之后每一列也都借助进位加到 $10$

Work from the right: the units digits must add to $10$ with a carry, and every other column then adds to $10$ with that carry

解答：

竖式相减得 $\begin{aligned} &1{,}000{,}000{,}000{,}000 \\ &\quad {}- 777{,}777{,}777{,}777 \\ &= 222{,}222{,}222{,}223 \end{aligned}\text{。}$

也可以想成：从 $777{,}777{,}777{,}777$ 增加到 $1{,}000{,}000{,}000{,}000$ ，个位要加 $3$ ，其余十一位各加 $2$ 。

所以正确答案是 B。

Lining up the subtraction gives $\begin{aligned} &1{,}000{,}000{,}000{,}000 \\ &\quad {}- 777{,}777{,}777{,}777 \\ &= 222{,}222{,}222{,}223. \end{aligned}$

Equivalently, to climb from $777{,}777{,}777{,}777$ up to $1{,}000{,}000{,}000{,}000$ you add $3$ in the units place and $2$ in each of the other eleven places.

Thus, the correct answer is B .

2.

下面算式的值是多少？ $\frac{16+8}{4-2}\text{？}$

What is the value of $\frac{16+8}{4-2}?$

$4$

$8$

$12$

$16$

$20$

答案：C

知识点：运算顺序整数运算

难度评级：450

小提示：

分数线表示整个分子除以整个分母

The fraction bar groups the whole numerator over the whole denominator

大提示：

先分别化简 $16+8$ 和 $4-2$ ，再相除

Simplify $16+8$ and $4-2$ separately before dividing

解答：

$\frac{16+8}{4-2} = \frac{24}{2} = 12\text{。}$

所以正确答案是 C。

$\frac{16+8}{4-2} = \frac{24}{2} = 12.$

Thus, the correct answer is C .

3.

二十万乘以二十万等于

Two hundred thousand times two hundred thousand equals

四十万

four hundred thousand

四百万

four million

四万

forty thousand

四亿

four hundred million

四百亿

forty billion

答案：E

知识点：位值指数

难度评级：450

小提示：

把二十万写成 $200{,}000$

Write two hundred thousand as $200{,}000$

大提示：

先乘前面的数字，再数零的个数：每个因数都有五个零

Multiply the leading digits, then count the zeros: each factor has five zeros

解答：

$200{,}000 \times 200{,}000$ $= 4 \times 10^{10}$ $= 40{,}000{,}000{,}000$ ，也就是四百亿。

所以正确答案是 E。

$200{,}000 \times 200{,}000$ $= 4 \times 10^{10}$ $= 40{,}000{,}000{,}000,$ which is forty billion.

Thus, the correct answer is E .

4.

如果 $991 + 993 + 995$ $+ 997 + 999$ $= 5000 - N$ ，那么 $N =$

If $991 + 993 + 995$ $+ 997 + 999$ $= 5000 - N,$ then $N =$

$5$

$10$

$15$

$20$

$25$

答案：E

知识点：等差数列

难度评级：560

小提示：

这五个数都只比 $1000$ 小一点

Each of the five numbers is just a little less than $1000$

大提示：

把每个数写成 $1000 - k$ ，总和就变成 $5000 - (9+7+5+3+1)$

Write each as $1000 - k;$ the sum becomes $5000 - (9+7+5+3+1)$

解答：

因为 $\begin{aligned} &991+993+995 \\ &\quad {}+997+999 \\ &= (1000-9)+(1000-7) \\ &\quad {}+(1000-5) \\ &\quad {}+(1000-3)+(1000-1) \\ &= 5000 - (9+7+5+3+1) \\ &= 5000 - 25 \end{aligned}\text{，}$ 所以 $N = 25$ 。

所以正确答案是 E。

Since $\begin{aligned} &991+993+995 \\ &\quad {}+997+999 \\ &= (1000-9)+(1000-7) \\ &\quad {}+(1000-5) \\ &\quad {}+(1000-3)+(1000-1) \\ &= 5000 - (9+7+5+3+1) \\ &= 5000 - 25, \end{aligned}$ we get $N = 25.$

Thus, the correct answer is E .

5.

一个“多米诺骨牌”由两个小正方形组成：

。下面哪一个“棋盘”不能被若干个不重叠的完整多米诺骨牌恰好完全覆盖？

A “domino” is made up of two small squares:

. Which of the “checkerboards” illustrated below CANNOT be covered exactly and completely by a whole number of non-overlapping dominoes?

$3 \times 4$

$3 \times 5$

$4 \times 4$

$4 \times 5$

$6 \times 3$

答案：B

知识点：铺砖奇偶性

难度评级：800

小提示：

每个多米诺骨牌恰好覆盖两个小正方形，所以先看每个棋盘有多少个小正方形

Each domino covers exactly two small squares, so focus on how many small squares each board has.

大提示：

棋盘要被完全覆盖，总格数必须是偶数；用两个维度相乘求每个棋盘的格数

A board can be fully covered only if its total number of squares is even; find the number of squares in each board by multiplying its two dimensions.

解答：

每个多米诺骨牌覆盖 $2$ 个小正方形，所以能被不重叠多米诺完全覆盖的棋盘必须有偶数个小正方形。

各棋盘的格数为： $3 \times 4 = 12$ ， $3 \times 5 = 15$ ， $4 \times 4 = 16$ ， $4 \times 5 = 20$ ， $6 \times 3 = 18$ 。只有 $3 \times 5 = 15$ 是奇数，所以不能被完全覆盖。（每个格数为偶数的棋盘都有一条边长是偶数，很容易用多米诺骨牌铺满。）

所以正确答案是 B。

Every domino covers exactly $2$ squares, so any board that is completely covered by non-overlapping dominoes must contain an even number of small squares.

Counting squares: $3 \times 4 = 12,$ $3 \times 5 = 15,$ $4 \times 4 = 16,$ $4 \times 5 = 20,$ and $6 \times 3 = 18.$ Only $3 \times 5 = 15$ is odd, so that board cannot be covered. (Each of the even boards has a side of even length and is easily tiled with dominoes.)

Thus, the correct answer is B .

6.

下面数组中的哪个数既是它所在列中最大的数，又是它所在行中最小的数？（列是上下方向，行是左右方向。）

$\begin{array}{ccccc} 10 & 6 & 4 & 3 & 2 \\ 11 & 7 & 14 & 10 & 8 \\ 8 & 3 & 4 & 5 & 9 \\ 13 & 4 & 15 & 12 & 1 \\ 8 & 2 & 5 & 9 & 3 \end{array}$

Which number in the array below is both the largest in its column and the smallest in its row? (Columns go up and down, rows go right and left.)

$\begin{array}{ccccc} 10 & 6 & 4 & 3 & 2 \\ 11 & 7 & 14 & 10 & 8 \\ 8 & 3 & 4 & 5 & 9 \\ 13 & 4 & 15 & 12 & 1 \\ 8 & 2 & 5 & 9 & 3 \end{array}$

$1$

$6$

$7$

$12$

$15$

答案：C

知识点：系统列举极端原理

难度评级：800

小提示：

先找出五列中每一列的最大数

First find the largest number in each of the five columns

大提示：

再检查这些列最大值中，哪一个也是它所在行的最小数

Then check which of those column-maxima is also the smallest number in its own row

解答：

五列的最大数依次为 $13$ （第 $1$ 列）、 $7$ （第 $2$ 列）、 $15$ （第 $3$ 列）、 $12$ （第 $4$ 列）和 $9$ （第 $5$ 列）。

这些数中，只有 $7$ 是它所在行中的最小数，因为第 $2$ 行是 $11$ 、 $7$ 、 $14$ 、 $10$ 、 $8$ 。

所以正确答案是 C。

The largest entry in each column is $13$ (column $1$ ), $7$ (column $2$ ), $15$ (column $3$ ), $12$ (column $4$ ), and $9$ (column $5$ ).

Of these, only $7$ is the smallest number in its own row (row $2$ is $11,$ $7,$ $14,$ $10,$ $8$ ).

Thus, the correct answer is C .

7.

下列表达式 $\tiny\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$ 的值最接近

The value of $\tiny\frac{(487{,}000)(12{,}027{,}300) + (9{,}621{,}001)(487{,}000)}{(19{,}367)(0.05)}$ is closest to

$10{,}000{,}000$

$100{,}000{,}000$

$1{,}000{,}000{,}000$

$10{,}000{,}000{,}000$

$100{,}000{,}000{,}000$

答案：D

知识点：估算分配律

难度评级：890

小提示：

分子有公因数 $487{,}000$

The numerator shares a common factor of $487{,}000$

大提示：

先把每个数四舍五入到一个主要数字再相除，例如 $487{,}000 \approx 500{,}000$ ，且 $(19{,}367)(0.05) \approx 1000$

Round each number to a single leading digit before dividing, for example $487{,}000 \approx 500{,}000$ and $(19{,}367)(0.05) \approx 1000$

解答：

分子可提取公因数： $487{,}000$ $\cdot\, (12{,}027{,}300 + 9{,}621{,}001)$ 。

粗略估计为 $500{,}000$ $\times (10{,}000{,}000 + 10{,}000{,}000)$ $= 500{,}000 \times 2$ $\times 10^{7}$ 。分母约为 $20{,}000 \times 0.05 = 1000$ 。

所以整个值约为 $\dfrac{500{,}000 \times 2 \times 10^{7}}{1000}$ $= 10^{10}$ $= 10{,}000{,}000{,}000$ 。

所以正确答案是 D。

Factor the numerator: $487{,}000$ $\cdot\, (12{,}027{,}300 + 9{,}621{,}001).$

Rounding to leading digits, this is about $500{,}000$ $\times (10{,}000{,}000 + 10{,}000{,}000)$ $= 500{,}000 \times 2$ $\times 10^{7}.$ The denominator is about $20{,}000 \times 0.05 = 1000.$

So the value is roughly $\dfrac{500{,}000 \times 2 \times 10^{7}}{1000}$ $= 10^{10}$ $= 10{,}000{,}000{,}000.$

Thus, the correct answer is D .

8.

从集合 $\{-24, -3, -2, 1, 2, 8\}$ 中选两个数相除，能形成的最大商是多少？

What is the largest quotient that can be formed using two numbers chosen from the set $\{-24, -3, -2, 1, 2, 8\}?$

$-24$

$-3$

$8$

$12$

$24$

答案：D

知识点：最优化分类讨论

难度评级：890

小提示：

最大的商应该是正数，并且数值尽可能大

A quotient is largest when it is positive and large in size

大提示：

负数除以负数是正数；试把绝对值最大的负数除以绝对值最小的负数

A negative divided by a negative is positive; try the largest-magnitude negative over the smallest-magnitude negative

解答：

要得到较大的商，应让商为正，所以可以用两个正数或两个负数。

最好的正数组合是 $\dfrac{8}{1} = 8$ 。最好的负数组合是 $\dfrac{-24}{-2} = 12$ 。较大的是 $12$ 。

所以正确答案是 D。

For a large quotient it should be positive, so use two positive numbers or two negative numbers.

Best positive pair: $\dfrac{8}{1} = 8.$ Best negative pair: $\dfrac{-24}{-2} = 12.$ The larger is $12.$

Thus, the correct answer is D .

9.

从 $1$ 到 $46$ 的整数中，有多少个能被 $3$ 或 $5$ 或两者都整除？

How many whole numbers from $1$ through $46$ are divisible by either $3$ or $5$ or both?

$18$

$21$

$24$

$25$

$27$

答案：B

知识点：容斥原理倍数

难度评级：890

小提示：

分别数出 $3$ 的倍数和 $5$ 的倍数

Count the multiples of $3$ and the multiples of $5$ separately

大提示：

同时能被 $3$ 和 $5$ 整除的数是 $15$ 的倍数，会被数两次，所以要减去一次

Numbers divisible by both $3$ and $5$ are multiples of $15$ and get counted twice, so subtract them once

解答：

不超过 $46$ 的 $3$ 的倍数有 $15$ 个， $5$ 的倍数有 $9$ 个。 $15$ 的倍数有 $3$ 个，即 $15$ 、 $30$ 、 $45$ ，它们被重复计算了。

用容斥原理，所求个数是 $15 + 9 - 3 = 21$ 。

所以正确答案是 B。

There are $15$ multiples of $3$ and $9$ multiples of $5$ up to $46.$ The $3$ multiples of $15$ (namely $15,$ $30,$ $45$ ) were counted twice.

By inclusion-exclusion, the count is $15 + 9 - 3 = 21.$

Thus, the correct answer is B .

10.

平行四边形 $ABCD$ 围成区域的面积是多少平方单位？

The area in square units of the region enclosed by parallelogram $ABCD$ is

$6$

$8$

$12$

$15$

$18$

答案：B

知识点：坐标几何平行四边形面积

难度评级：860

小提示：

取 $\overline{BC}$ 为底；它是从 $B(0,2)$ 到 $C(4,2)$ 的水平线段

Take $\overline{BC}$ as the base; it is horizontal from $B(0,2)$ to $C(4,2)$

大提示：

面积等于底乘高，其中高是两条水平边之间的竖直距离

The area is base times height, where the height is the vertical distance between the two horizontal sides

解答：

边 $BC$ 从 $(0,2)$ 到 $(4,2)$ ，所以底长为 $4$ 。相对边 $AD$ 在 $x$ 轴上，所以高为 $2$ 。

面积是 $4 \times 2 = 8$ 。

所以正确答案是 B。

Side $BC$ runs from $(0,2)$ to $(4,2),$ so the base is $4.$ The opposite side $AD$ lies on the $x$ -axis, so the height is $2.$

The area is $4 \times 2 = 8.$

Thus, the correct answer is B .

11.

从 $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ 中可以选出若干组三个不同的数，它们的和为 $15$ 。其中有多少组包含 $5$ ？

There are several sets of three different numbers whose sum is $15$ which can be chosen from $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}.$ How many of these sets contain a $5?$

$3$

$4$

$5$

$6$

$7$

答案：B

知识点：数对计数系统列举

难度评级：890

小提示：

如果 $5$ 是三个数之一，那么另外两个数的和必须是 $10$

If $5$ is one of the three numbers, the other two must add to $10$

大提示：

列出从 $\{1,2,3,4,6,7,8,9\}$ 中选出的、和为 $10$ 的不同数对

List the pairs of different numbers from $\{1,2,3,4,6,7,8,9\}$ that sum to $10$

解答：

选定 $5$ 后，另外两个不同的数必须和为 $10$ 。这样的数对是 $1+9$ 、 $2+8$ 、 $3+7$ 、 $4+6$ ，共 $4$ 组。

所以正确答案是 B。

With $5$ chosen, the other two different numbers must sum to $10.$ The pairs are $1+9,$ $2+8,$ $3+7,$ $4+6,$ giving $4$ sets.

Thus, the correct answer is B .

12.

如果 $\begin{aligned} &\frac{2+3+4}{3} \\ &= \frac{1990+1991+1992}{N} \end{aligned}\text{，}$ 那么 $N =$

If $\begin{aligned} &\frac{2+3+4}{3} \\ &= \frac{1990+1991+1992}{N}, \end{aligned}$ then $N =$

$3$

$6$

$1990$

$1991$

$1992$

答案：D

知识点：等差数列平均数

难度评级：820

小提示：

先化简左边

Simplify the left side first

大提示：

三个连续整数的和除以中间那个数，结果为 $3$

The sum of three consecutive integers divided by the middle one equals $3$

解答：

左边是 $\dfrac{9}{3} = 3$ 。右边是 $\dfrac{5973}{N}$ ，令它等于 $3$ ，得到 $N = 1991$ 。

也可以用规律： $(k-1) + k + (k+1) = 3k$ ，所以除以 $3$ 后得到中间项。这里中间项是 $1991$ 。

所以正确答案是 D。

The left side is $\dfrac{9}{3} = 3.$ The right side is $\dfrac{5973}{N},$ and setting it equal to $3$ gives $N = 1991.$

Equivalently, $(k-1) + k + (k+1) = 3k,$ so dividing by $3$ leaves the middle term. Here the middle term is $1991.$

Thus, the correct answer is D .

13.

下列乘积末尾有多少个零？ $\begin{aligned} &25 \times 25 \times 25 \times 25 \\ &\quad {}\times 25 \times 25 \times 25 \\ &\quad {}\times 8 \times 8 \times 8\text{？} \end{aligned}$

How many zeros are at the end of the product $\begin{aligned} &25 \times 25 \times 25 \times 25 \\ &\quad {}\times 25 \times 25 \times 25 \\ &\quad {}\times 8 \times 8 \times 8? \end{aligned}$

$3$

$6$

$9$

$10$

$12$

答案：C

知识点：末尾零质因数分解

难度评级：1000

小提示：

一个末尾零来自一个因数 $10 = 2 \times 5$

A trailing zero comes from a factor of $10 = 2 \times 5$

大提示：

分别数出因数 $2$ 和因数 $5$ 的总数；末尾零的个数是两者中较小的那个

Count the total factors of $2$ and of $5;$ the number of trailing zeros is the smaller count

解答：

因为 $25 = 5^2$ ，七个 $25$ 提供 $5^{14}$ 。因为 $8 = 2^3$ ，三个 $8$ 提供 $2^9$ 。

末尾零的个数是 $\min(14, 9) = 9$ 。

所以正确答案是 C。

Since $25 = 5^2,$ the seven $25$ ’s give $5^{14}.$ Since $8 = 2^3,$ the three $8$ ’s give $2^9.$

The number of trailing zeros is $\min(14, 9) = 9.$

Thus, the correct answer is C .

14.

几名学生参加三场赛跑。每场比赛第一名得 $5$ 分，第二名得 $3$ 分，第三名得 $1$ 分。没有并列名次。一个学生在三场比赛中至少要得到多少分，才能保证他的总分高于任何其他学生？

Several students are competing in a series of three races. A student earns $5$ points for winning a race, $3$ points for finishing second, and $1$ point for finishing third. There are no ties. What is the smallest number of points that a student must earn in the three races to be guaranteed of earning more points than any other student?

$9$

$10$

$11$

$13$

$15$

答案：D

知识点：极端原理

难度评级：1090

小提示：

要保证领先，要考虑最坏情况：另一个学生也尽可能得高分

To be guaranteed the lead, plan for the worst case where another student does as well as possible

大提示：

依次检查可能的三场总分；对每个总分，判断剩余名次是否允许另一名学生追平或超过它

Check the attainable three-race totals in increasing order; for each one, see whether the remaining places allow a rival to tie or exceed it

解答：

总分 $11$ 并不能保证第一，例如 $5+5+1$ 可能被另一个学生也得到 $11$ 分。

但如果一个学生得到 $5+5+3 = 13$ 分，那么剩下的名次使任何其他学生最多得到 $3+3+5 = 11$ 分。因此 $13$ 分能保证领先。

所以正确答案是 D。

A total of $11$ (for example $5+5+1$ ) does not guarantee first place, since another student could also reach $11.$

But if one student scores $5+5+3 = 13,$ the remaining places give every other student at most $3+3+5 = 11.$ So $13$ points guarantees the lead.

Thus, the correct answer is D .

15.

一个长方体的六个面都是长方形。如图，从这个长方体中切去一个边长为一英尺的立方体。新立体的表面积比原立体的表面积多或少多少平方英尺？

All six sides of a rectangular solid were rectangles. A one-foot cube was cut out of the rectangular solid as shown. The total number of square feet in the surface of the new solid is how many more or less than that of the original solid?

少 $2$

$2$ less

少 $1$

$1$ less

相同

the same

多 $1$

$1$ more

多 $2$

$2$ more

答案：C

知识点：表面积立体几何

难度评级：1140

小提示：

切掉立方体会去掉一些原本在表面的面，也会露出一些新的面；比较这两个面积

Cutting the cube removes some faces that were on the surface but uncovers new ones; compare the two amounts

大提示：

被切掉的小立方体的上、前、后三个面原本在表面上，去掉 $3$ 平方英尺；凹口的底面和两个侧壁成为新表面，又增加 $3$ 平方英尺

The cube’s top, front, and back faces were on the surface ( $3$ square feet removed); the notch’s floor and its two side walls become new surface ( $3$ square feet added)

解答：

被切掉的小立方体有三个面原来在立体表面上：上面、前面、后面，所以移除了 $3$ 平方英尺表面积。

切出凹口后，又露出三个新面：凹口底面和两个侧壁，共增加 $3$ 平方英尺。因此表面积不变。

所以正确答案是 C。

The removed cube had three faces on the surface of the solid (top, front, and back), so $3$ square feet of surface are removed.

Cutting it out exposes three new faces (the floor of the notch and its two side walls), adding $3$ square feet. The surface area is unchanged.

Thus, the correct answer is C .

16.

一张纸上的 $16$ 个小正方形按图编号。纸平放在桌上，按下面顺序对折四次：

$(1)$ 将上半部分折到下半部分上；

$(2)$ 将下半部分折到上半部分上；

$(3)$ 将右半部分折到左半部分上；

$(4)$ 将左半部分折到右半部分上。

第 $4$ 步之后，哪个编号的小正方形在最上面？

$\begin{array}{|c|c|c|c|} \hline 1 & 2 & 3 & 4 \\ \hline 5 & 6 & 7 & 8 \\ \hline 9 & 10 & 11 & 12 \\ \hline 13 & 14 & 15 & 16 \\ \hline \end{array}$

The $16$ squares on a piece of paper are numbered as shown in the diagram. While lying on a table, the paper is folded in half four times in the following sequence:

$(1)$ fold the top half over the bottom half;

$(2)$ fold the bottom half over the top half;

$(3)$ fold the right half over the left half;

$(4)$ fold the left half over the right half.

Which numbered square is on top after step $4?$

$\begin{array}{|c|c|c|c|} \hline 1 & 2 & 3 & 4 \\ \hline 5 & 6 & 7 & 8 \\ \hline 9 & 10 & 11 & 12 \\ \hline 13 & 14 & 15 & 16 \\ \hline \end{array}$

$1$

$9$

$10$

$14$

$16$

答案：B

知识点：折纸过程模拟

难度评级：1180

小提示：

每次折叠后，追踪哪些小正方形在纸叠的底部或顶部

After each fold, track which squares end up on the bottom (or top) of the stack

大提示：

第 $1$ 、 $2$ 次折叠留下中间两行，第 $3$ 、 $4$ 次折叠留下中间两列；最后一次左折到右会把 $9$ 号方格翻到最上面

Folds $1$ and $2$ leave the middle rows; folds $3$ and $4$ leave the middle columns, and the final left-over-right fold flips square $9$ to the top

解答：

第 $1$ 次上半折到下半后， $9\text{–}16$ 在底部。第 $2$ 次下半折到上半后， $9\text{–}12$ 在底部。第 $3$ 次右半折到左半后， $9$ 和 $10$ 在底部。

第 $4$ 次左半折到右半，会让 $10$ 在底部，并把 $9$ 带到最上面。

所以正确答案是 B。

Fold $1$ (top over bottom) leaves squares $9\text{–}16$ on the bottom. Fold $2$ (bottom over top) leaves $9\text{–}12$ on the bottom. Fold $3$ (right over left) leaves $9$ and $10$ on the bottom.

Fold $4$ (left over right) puts $10$ on the bottom and brings $9$ to the top.

Thus, the correct answer is B .

17.

一个礼堂有 $20$ 排座位，第一排有 $10$ 个座位。之后每一排都比前一排多一个座位。如果参加考试的学生可以坐在任何一排，但同一排中不能与另一个学生相邻，那么最多能安排多少名学生考试？

An auditorium with $20$ rows of seats has $10$ seats in the first row. Each successive row has one more seat than the previous row. If students taking an exam are permitted to sit in any row, but not next to another student in that row, then the maximum number of students that can be seated for an exam is

$150$

$180$

$200$

$400$

$460$

答案：C

知识点：有限制的排列等差数列取整函数

难度评级：1140

小提示：

一排有 $n$ 个座位时，若不能相邻，最多可坐 $\left\lceil \frac{n}{2} \right\rceil$ 人

In a row of $n$ seats, the most students with no two adjacent is $\left\lceil \frac{n}{2} \right\rceil$

大提示：

第 $k$ 排有 $9+k$ 个座位；将 $\left\lceil \frac{9+k}{2} \right\rceil$ 在前 $20$ 排中相加

Row $k$ has $9+k$ seats; add $\left\lceil \frac{9+k}{2} \right\rceil$ over the $20$ rows

解答：

第 $k$ 排有 $9+k$ 个座位，所以最多坐 $\left\lceil \frac{9+k}{2} \right\rceil$ 人。第 $1$ 到第 $20$ 排的座位数从 $10$ 到 $29$ ，对应最大人数为 $5$ 、 $6$ 、 $6$ 、 $7$ 、 $7$ 、 $8$ 、 $8$ 、 $\ldots$ 、 $14$ 、 $14$ 、 $15$ 。

这些数相加为 $200$ 。

所以正确答案是 C。

Row $k$ has $9+k$ seats, so it holds $\left\lceil \frac{9+k}{2} \right\rceil$ students. For rows $1$ through $20$ (seats $10$ through $29$ ) the maxima are $5,$ $6,$ $6,$ $7,$ $7,$ $8,$ $8,$ $\ldots,$ $14,$ $14,$ $15.$

These sum to $200.$

Thus, the correct answer is C .

18.

纵轴表示员工人数，但这张图的刻度被意外漏掉了。高斯公司有百分之几的员工已经在那里工作了 $5$ 年或更久？

The vertical axis indicates the number of employees, but the scale was accidentally omitted from this graph. What percent of the employees at the Gauss Company have worked there for $5$ years or more?

$9\%$

$23\dfrac13\%$

$30\%$

$42\dfrac67\%$

$50\%$

答案：C

知识点：数据与图表解读百分数

难度评级：890

小提示：

因为纵轴刻度未知，所以不要用具体数值，而是数 X 的个数

Since the vertical scale is unknown, work with the number of X’s rather than any value

大提示：

用工作年数 $5$ 到 $10$ 上方的 X 个数除以所有 X 的个数

Divide the X’s over years $5$ through $10$ by the total number of X’s

解答：

不论缺失的刻度是多少，每个 X 都代表相同数量的员工。共有 $9$ 个 X 分布在工作年数 $5$ 到 $10$ 的各栏上，全图共有 $30$ 个 X。

所以比例是 $\dfrac{9}{30} = 30\%$ 。

所以正确答案是 C。

No matter the missing scale, each X represents the same number of employees. There are $9$ X’s over years $5$ through $10,$ and $30$ X’s in all.

So the fraction is $\dfrac{9}{30} = 30\%.$

Thus, the correct answer is C .

19.

$10$ 个不同正整数的平均数是 $10$ 。这些数中任意一个可能取得的最大值是

The average (arithmetic mean) of $10$ different positive whole numbers is $10.$ The largest possible value of any of these numbers is

$10$

$50$

$55$

$90$

$91$

答案：C

知识点：平均数极端原理

难度评级：1030

小提示：

这十个数的总和必须是 $10 \times 10 = 100$

The ten numbers must sum to $10 \times 10 = 100$

大提示：

要让其中一个数尽可能大，就让另外九个不同正整数尽可能小，即取 $1, 2, \ldots, 9$ 这九个数

To make one number as large as possible, make the other nine as small as possible: $1, 2, \ldots, 9$

解答：

这十个数的总和是 $100$ 。要使其中一个数最大，另外九个不同正整数应尽可能小，即 $1 + 2 + \cdots + 9 = 45$ 。

最大的那个数就是 $100 - 45 = 55$ 。

所以正确答案是 C。

The ten numbers sum to $100.$ To maximize one of them, the other nine (all different positive whole numbers) should be as small as possible: $1 + 2 + \cdots + 9 = 45.$

The largest number is then $100 - 45 = 55.$

Thus, the correct answer is C .

20.

在所示加法算式中，每个数字都被一个字母代替。若不同字母表示不同数字，那么 $C =$

$\begin{array}{cccc} & A & B & C \\ & & A & B \\ + & & & A \\ \hline & 3 & 0 & 0 \end{array}$

In the addition problem shown, each digit has been replaced by a letter. If different letters represent different digits, then $C =$

$\begin{array}{cccc} & A & B & C \\ & & A & B \\ + & & & A \\ \hline & 3 & 0 & 0 \end{array}$

$1$

$3$

$5$

$7$

$9$

答案：A

知识点：数字谜位值

难度评级：1140

小提示：

用 $A, B, C$ 写出这个加法和

Write the sum in terms of $A, B, C$

大提示：

把三个数相加，得到 $100A + 10B + C$ $+ 10A + B + A$ $= 111A + 11B + C$ $= 300$ 这个等式

$100A + 10B + C$ $+ 10A + B + A$ $= 111A + 11B + C$ $= 300$

解答：

三个数相加为 $111A + 11B + C = 300$ 。若 $A = 1$ ，数值太小；若 $A \ge 3$ ，数值太大，所以 $A = 2$ 。

此时 $11B + C = 78$ ，只能有 $B = 7$ 且 $C = 1$ 。所以 $C = 1$ 。

所以正确答案是 A。

The three numbers add to $111A + 11B + C = 300.$ Since $A = 1$ is too small and $A \ge 3$ is too large, $A = 2.$

Then $11B + C = 78,$ which forces $B = 7$ and $C = 1.$ So $C = 1.$

Thus, the correct answer is A .

21.

温度每升高 $3^\circ$ ，某种气体的体积就膨胀 $4$ 立方厘米。如果温度为 $32^\circ$ 时气体体积是 $24$ 立方厘米，那么温度为 $20^\circ$ 时气体体积是多少立方厘米？

For every $3^\circ$ rise in temperature, the volume of a certain gas expands by $4$ cubic centimeters. If the volume of the gas is $24$ cubic centimeters when the temperature is $32^\circ,$ what was the volume of the gas in cubic centimeters when the temperature was $20^\circ?$

$8$

$12$

$15$

$16$

$40$

答案：A

知识点：速率逆推法

难度评级：950

小提示：

温度从 $32^\circ$ 降到 $20^\circ$ ，变化了 $12^\circ$

The temperature dropped from $32^\circ$ to $20^\circ,$ a change of $12^\circ$

大提示：

每下降 $3^\circ$ ，体积减少 $4$ 立方厘米

Each $3^\circ$ drop shrinks the volume by $4$ cubic centimeters

解答：

从 $32^\circ$ 到 $20^\circ$ 是下降 $12^\circ$ ，也就是 $4$ 个 $3^\circ$ 。

体积减少 $4 \times 4 = 16$ 立方厘米，所以从 $24$ 降到 $24 - 16 = 8$ 。

所以正确答案是 A。

From $32^\circ$ to $20^\circ$ is a $12^\circ$ decrease, which is $4$ steps of $3^\circ.$

The volume decreases by $4 \times 4 = 16$ cubic centimeters, from $24$ down to $24 - 16 = 8.$

Thus, the correct answer is A .

22.

每个转盘都被平均分成 $3$ 部分。转动两个转盘，把得到的两个数相乘。乘积为偶数的概率是多少？

Each spinner is divided into $3$ equal parts. The results obtained from spinning the two spinners are multiplied. What is the probability that this product is an even number?

$\dfrac13$

$\dfrac12$

$\dfrac23$

$\dfrac79$

$1$

答案：D

知识点：对立事件概率奇偶性

难度评级：1000

小提示：

乘积只有在两个数都是奇数时才是奇数；除此以外都是偶数

A product is even unless both spun numbers are odd

大提示：

先求奇数乘积的概率，也就是奇数乘奇数，再用 $1$ 减去它

Find the probability of an odd product (odd times odd), then subtract from $1$

解答：

乘积为奇数当且仅当两个数都是奇数。第一个转盘转到奇数 $1$ 或 $3$ 的概率是 $\dfrac23$ ，第二个转盘转到奇数 $5$ 的概率是 $\dfrac13$ 。

所以乘积为奇数的概率是 $\dfrac23 \times \dfrac13 = \dfrac29$ ，乘积为偶数的概率是 $1 - \dfrac29 = \dfrac79$ 。

所以正确答案是 D。

The product is odd only when both numbers are odd. The first spinner is odd ( $1$ or $3$ ) with probability $\dfrac23,$ and the second is odd ( $5$ ) with probability $\dfrac13.$

So the product is odd with probability $\dfrac23 \times \dfrac13 = \dfrac29,$ and even with probability $1 - \dfrac29 = \dfrac79.$

Thus, the correct answer is D .

23.

毕达哥拉斯高中的乐队有 $100$ 名女生和 $80$ 名男生。毕达哥拉斯高中的管弦乐队有 $80$ 名女生和 $100$ 名男生。有 $60$ 名女生同时属于乐队和管弦乐队。总共有 $230$ 名学生属于乐队或管弦乐队或两者都属于。属于乐队但不属于管弦乐队的男生人数是

The Pythagoras High School band has $100$ female and $80$ male members. The Pythagoras High School orchestra has $80$ female and $100$ male members. There are $60$ females who are members in both band and orchestra. Altogether, there are $230$ students who are in either band or orchestra or both. The number of males in the band who are NOT in the orchestra is

$10$

$20$

$30$

$50$

$70$

答案：A

知识点：容斥原理韦恩图

难度评级：1200

小提示：

先用 $100 + 80 - 60$ 求属于至少一个团体的女生人数

First find how many females are in at least one group, using $100 + 80 - 60$

大提示：

用总人数 $230$ 减去女生人数，得到属于至少一个团体的男生人数；再用容斥原理求同时属于两个团体的男生人数

Subtract the females from the $230$ total to get the males in at least one group, then use inclusion–exclusion to find how many males are in both groups

解答：

属于乐队或管弦乐队的女生人数为 $100 + 80 - 60 = 120$ 。所以属于至少一个团体的男生人数为 $230 - 120 = 110$ 。

乐队有 $80$ 名男生，管弦乐队有 $100$ 名男生，因此两个团体都有的男生人数是 $80 + 100 - 110 = 70$ 。属于乐队但不属于管弦乐队的男生有 $80 - 70 = 10$ 人。

所以正确答案是 A。

Females in band or orchestra: $100 + 80 - 60 = 120.$ So males in at least one group: $230 - 120 = 110.$

With $80$ males in band and $100$ in orchestra, the males in both are $80 + 100 - 110 = 70.$ Hence males in band but not orchestra: $80 - 70 = 10.$

Thus, the correct answer is A .

24.

一个边长为 $3$ 厘米的正方体被切成 $N$ 个较小的正方体，且这些小正方体不全同样大小。如果每个小正方体的边长都是整厘米数，那么 $N =$

A cube of edge $3$ cm is cut into $N$ smaller cubes, not all the same size. If the edge of each of the smaller cubes is a whole number of centimeters, then $N =$

$4$

$8$

$12$

$16$

$20$

答案：E

知识点：正方体体积

难度评级：1140

小提示：

小正方体的边长必须是整数，所以只能是边长 $1$ 或边长 $2$

The smaller cubes must have whole-number edges, so each is edge $1$ or edge $2$

大提示：

只能放下一个边长为 $2$ 的正方体；剩下的体积 $27 - 8$ 用单位正方体填满

Only one edge- $2$ cube fits; the rest are unit cubes filling the remaining volume $27 - 8$

解答：

这个 $3 \times 3 \times 3$ 正方体的体积是 $27$ 。由于小正方体更小且边长为整数，它们的边长只能是 $1$ 或 $2$ 。原正方体内不能放下两个互不相交的边长为 $2$ 的正方体，因为要将它们分开，某个方向上的总跨度至少为 $4$ 。因此最多只能放下一个边长为 $2$ 的正方体。

剩余体积 $27 - 8 = 19$ ，由 $19$ 个单位正方体填满。总数是 $1 + 19 = 20$ 个正方体。

所以正确答案是 E。

The $3 \times 3 \times 3$ cube has volume $27.$ Since the cubes are smaller and have whole-number edge lengths, their edges are $1$ or $2$ . Two edge- $2$ cubes cannot be disjoint inside the original cube: separating them would require a total span of at least $4$ in some direction. Thus at most one edge- $2$ cube fits.

The remaining $27 - 8 = 19$ of volume is filled by $19$ unit cubes. That is $1 + 19 = 20$ cubes.

Thus, the correct answer is E .

25.

一个等边三角形最初被涂成黑色。每次改变时，每个黑色三角形的中间四分之一变成白色。经过五次改变后，仍为黑色的部分占原三角形面积的几分之几？

An equilateral triangle is originally painted black. Each time the triangle is changed, the middle fourth of each black triangle turns white. After five changes, what fractional part of the original area of the black triangle remains black?

$\dfrac{1}{1024}$

$\dfrac{15}{64}$

$\dfrac{243}{1024}$

$\dfrac14$

$\dfrac{81}{256}$

答案：C

知识点：等比数列面积比

难度评级：1140

小提示：

改变一次后，中间四分之一变白，所以黑色面积剩下 $\dfrac34$

After one change the middle fourth turns white, so $\dfrac34$ of the black area remains

大提示：

每次改变都会把黑色面积乘以 $\dfrac34$ ，所以五次后乘以 $\left(\dfrac34\right)^5$

Each change multiplies the black area by $\dfrac34,$ so after five changes multiply by $\left(\dfrac34\right)^5$

解答：

每一次改变都会让当前黑色面积的 $\dfrac34$ 保持黑色。五次改变后，黑色面积比例为 $\left(\dfrac34\right)^5 = \dfrac{243}{1024}\text{。}$

所以正确答案是 C。

Each change leaves $\dfrac34$ of the current black area black. After five changes the black fraction is $\left(\dfrac34\right)^5 = \dfrac{243}{1024}.$

Thus, the correct answer is C .